The cardinality of a set of increasing rational sequences.

Question

For any $a\in(0,1)$, set $$E(a)=\{(r_1,r_2,\cdots,r_n\cdots)\,|\,r_n\in\mathbb{Q}\cap[0,1], r_n\leq r_{n+1},\forall n\geq1, \sum_{n=1}^\infty\frac{r_n}{2^n}=a\}.$$ Determine the cardinal number of $E(a)$, $|E(a)|$.

Answer 1

Let $\{q_n\}$ be any sequence of positive rational numbers. Let $a_n=\frac {q_n} {nN2^{n}(1+q_n)}$ where the integer $N$ is so large that $\sum na_n <a/2$. Next take a sequence $\{s_n\}$ of positive rationals such that $\sum (2n-1)s_n =a-2\sum na_n$. [see details below]. Let $t_{2n}=a_n$ and $t_{2n-1}=s_n$. Then $\sum nt_n=a$. Now let $r_n=t_1+t_2+...+t_n$. Then $\{r_n\} \in E(a)$. We have produced a one-to-one map from the set of all positive sequences of rationals into the set $E(a)$. The cardinality of $E(a)$ is therefore $c$.

We have used the fact that given any $b>0$ there exists a sequence of positive rationals $\{s_n\}$ with $\sum (2n-1)s_n=b$. To see this take a sequence $\{d_n\}$ of non-negative rationals increasing to $b$ with $d_1=0$. Let $s_n=\frac {d_{n+1}-d_n} {2n-1}$. Then the n-th partial sum of the series $\sum (2n-1)s_n=b$ is $d_{n+1}-d_1=d_{n+1}$ which tends to $b$.